Bartle - Introduction To Real Analysis - Chapter 6 Solutions

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Bartle - Introduction to Real Analysis - Chapter 6 SolutionsSection 6.2Problem 6.2-4. Let a1 , a2 , . . . , an be real numbers and let f be defined on R byf (x) nX(ai x)2 f or x R.i 0Find the unique point of relative minimum for f .Solution: The first derivative of f is:f 0 (x) 2nX(ai x).i 10Equating f to zero, we find the relative extrema c R as follows:"#nnXXf 0 (c) 2(ai c) 2 nc ai 0.i 1i 1Accordingly, the only relative extremum of f on R is:n1Xai ,c n i 1or more compactly the mean of the constants ai . By the Interior-Extremum Theorem, this extremum is unique (andtherefore global) because there are no other points in R where f 0 vanishes and the interval on which f is unbounded.We can use to First Derivative Test for Extrema to determine if f is at a minimum or maximum at c. For δ 0, wehave for for x c δ c:f 0 (x) f (c δ) 2nnXX(ai c δ) f 0 (c) 2( δ) 2δ 0,i 1i 1where we have relied on f 0 (c) 0. For x c δ c:f 0 (x) f (c δ) 2nX(ai c δ) f 0 (c) 2i 1nXδ 2δ 0.i 1By the First Derivative Test for Extrema, f is at a minimum at c.Problem 6.2-6. Use the Mean Value Theorem to prove that sin x sin y x y .Solution: Let f (x) sin x, so f 0 (x) cos x. For all x R, we have 1 cos x 1. Let x, y R. We may assumewithout loss of generality that x y. By the Mean Value Theorem, there is a c (x, y) such that:sin x sin y (x y)f 0 (c) (x y) cos c.Because cos c 1, it follows that sin x sin y x y .Problem 6.2-9. Let f : R R be defined by f (x) : 2x4 x4 sin(1/x) for x 6 0 and f (0) : 0. Show that f has an absoluteminimum at x 0, but that its derivative has both positive and negative values in every neighborhood of 0.Solution: By the definition of the derivative:2x4 x4 sin(1/x) lim [2x3 x3 sin(1/x)] 0.x 0x 0x 0f 0 (0) lim

By the Interior Extremum Theorem, there is a relative extremum at x 0. Because sin x 1 for all x R, it followsthat f (x) x4 [2 sin(1/x)] x4 (2 1) 0. Therefore, x 0 is an absolute minimum on R.For x 6 0, we have: 11f 0 (x) 8x3 3x3 sin x2 cos.xxLet δ 0 and Vδ (0) be a δ-neighborhood of zero. By the Archimedean Property, there is an n0 N such that0 1/n0 2πδ. It follows that there is an n N where n n0 1 such that 0 1/n 2πδ. Accordingly, lettingx0 1/(2πn), we see that x0 Vδ (0) and:14 πn8 0 2 2 0,8π 3 n34π n4π 3 n3where, in the last step, we relied upon n n0 1 1.By a similar argument, there is an n N such that 0 1/(n 1/4) 1/n 2πδ. Letting x00 1/[2π(n 1/4)], itfollows that x00 Vδ (0) and:f 0 (x0 ) f 0 (x00 ) 3118 3 0 0.8π 3 (n 1/4)38π (n 1/4)38π 3 (n 1/4)3Since δ is arbitrary, we conclude that f 0 on any neighborhood of zero attains both positive and negative values.Problem 6.2-11. Give an example of a uniformly continuous function on [0, 1] that is differentiable on (0, 1) but whosederivative is not bounded on (0, 1). Solution: Suppose f : [0, 1] R and f (x) x. If c [0, 1], then limx c x c f (c); therefore, f is continuouson [0, 1]. By the Uniform Continuity Theorem, f is uniformly continuous on [0, 1]. On (0, 1), the derivative of f is f 0 (x) 1/(2 x). Assumethat f 0 is boundedby M 0 on (0, 1). Then 1/(2 x) Mp for any x (0, 1). However, if x 1/(4M 2 ), then 1/(2 x) 1/(2 (1/4M 2 )) M , contradicting our assumptionthat f 0 is bounded. It follows that f 0 is unbounded on (0, 1).Problem 6.2-13. Let I be an interval and let f : I R be differentiable on I. Show that if f 0 is positive on I, then f isstrictly increasing on I.Solution: Suppose f 0 (x) 0 for all x I. Let x0 , x1 I where x0 x1 . Because f is differentiable on I by hypothesis(and therefore is continuous on I by Theorem 6.1.2), it follows from the Mean Value Theorem that there is a c I suchthat:f (x1 ) f (x0 ) f 0 (c)(x1 x0 )Because f 0 (c) 0 and x1 x0 0, the difference f (x1 ) f (x0 ) must be positive. Since x0 x1 are arbitrary pointsin I, the function f is strictly increasing on I.Problem 6.2-14. Let I be an interval and let f : I R be differentiable on I. Show that if f 0 is never 0 on I, then eitherf 0 (x) 0 for all x I or f 0 (x) 0 for all x I.Solution: Suppose f 0 (x) 6 0 for all x I. Now assume that there are x0 , x1 I where f 0 (x0 ) 0 and f 0 (x1 ) 0.Because f 0 (x0 ) 0 f 0 (x1 ), by Darboux’s Theorem, there is a c I such f 0 (c) 0, which contradicts our hypothesisthat f 0 is never zero on I. Therefore, either f 0 (x) 0 for all x I or f 0 (x) 0 for all x I.Problem 6.2-16. Let f : [0, ) R be differentiable on (0, ) and assume f 0 (x) b as x . (a) Show that for anyh 0, we have limx (f (x h) f (x))/h b. (b) Show that if f (x) a and x , then b 0. (c) Show thatlimx (f (x)/x) b.Page 2

Solution: Part (a) Because f is differentiable on (0, ), it is continuous on (0, ). For x, y in(0, ) where y x hfor any h 0, the Mean Value Theorem ensures that there is a cx (x, x h) such that:f (x h) f (x) f 0 (cx )(x h x) hf 0 (cx ).Accordingly, (f (x h) f (x))/h f 0 (cx ).By hypothesis, limx f 0 (x) b. Therefore, for any 0, there is a K( ) 0 such that if x K( ), then0 f (x) b . For any such x, it follows that cx x K( ). Consequently, for any x K( ): f 0 (cx ) b f (x h) f (x) b ,hfrom which it follows that limx (f (x h) f (x))/h) b.Part (b) If limx f (x) a, then for any 0, there is a K 0 ( ) such that if x K 0 ( ), then f (x) a . Becausex h x, it follows that f (x h) a , and therefore, limx f (x h) a. Consequently, limx [f (x h) f (x)] a a 0. It is then obvious that:f (x h) f (x)1 (0) 0.hhPart (c) [****IGNORE, NOT VALID SOLUTION***] It suffices to show that for 0, there is a K 00 ( ) 0 suchthat if x sup{0, K 00 ( )}, then:b limx f (2x) b ,(1)where the argument of f is permissible because x 0 ensures that 2x x.From part (a) with h x, we infer that for 0 b, there is a K( ) 0 such that if x sup{0, K( )}, then:f (2x) f (x) b .xNote that the upper bound on is acceptable because we are only concerned about small . As a result: f (2x) f (x) b ,xfrom which we have:f (x)f (x)f (2x) b***THIS STEP IS ERRONEOUS*** Substituting this inequality into equation (1) and letting K 00 ( ) sup{0, 1/2K( )},we have, for 0 and x K 00 ( ):f (2x)f (x) b b .xxConsequently, limx f (x)/x b, as we sought to prove.Problem 6.2-17. Let f, g be differentiable on R and suppose that f (0) g(0) and f 0 (x) g 0 (x) for all x 0. Show thatf (x) g(x) for all x 0.Solution: Let φ(x) g(x) f (x). Therefore, φ0 (x) g 0 (x) f 0 (x). Because f 0 (x) g 0 (x) for all x 0, it follows thatφ0 (x) 0 for all x 0. By Theorem 6.2.7, φ0 is increasing on [0, ). Therefore, if x0 , x1 [0, ) and x0 x1 , thenφ(x0 ) φ(x1 ). Since φ(0) 0, it follows that if x 0, then φ(x) g(x) f (x) 0 and therefore f (x) g(x).Problem 6.2-18. Let I : [a, b] and let f : I R be differentiable at c I. Show that for every 0 there exists δ 0such that if 0 x y δ and a x c y b, thenf (x) f (y) f 0 (c) x yPage 3

Solution: First note that x 6 y because, by hypothesis, x y 0. Because f is differentiable at c, for /2 0, we haveδ 0 0 where for x [a, c) and 0 x c δ 0 :f (x) f (c) f 0 (c) ,x c2(2)f (y) f (c) f 0 (c) .y c2(3)and, for y (c, b] where 0 y c δ 0 :Because x c and y c and x 6 y by hypothesis, it follows that x y x c and x y y c . Consequently,for equation (2), we have: 11 f (x) f (c) (x c)f 0 (c) f (c) f (x) (x c)f 0 (c) .2 x c x y (4)Similarly, for equation (3), we have: 11 f (y) f (c) (y c)f 0 (c) f (y) f (c) (y c)f 0 (c) .2 y c x y (5)Taking the sum of equations (4) and (5) and using the Triangle Inequality, we have: 1( f (y) f (c) (y c)f 0 (c) f (c) f (x) (x c)f 0 (c)) x y 1 f (y) f (x) f (c) f (c) yf 0 (c) xf 0 (c) cf 0 (c) cf 0 (c) x y 1f (x) f (y) f (y) f (x) f 0 (c)(x y) f 0 (c) , x y x y(6)which is true for 0 x c δ 0 and 0 y c δ 0 . From these relationships, we infer that 0 x y 2δ 0 . If we letδ 2δ 0 , we conclude that for any 0, the relationship in (6) holds for any 0 x y δ where a x c y b.Section 6.3Problem 6.3-4. Let f (x) : x2 for x rational and let f (x) : 0 for x irrational, and let g(x) sin x for x R. Use Theorem6.3.1 to show that limx 0 f (x)/g(x) 0. Explain why Theorem 6.3.3 cannot be used.Solution: The desired limit is in 0/0 indeterminate form. Theorem 6.3.1 requires both f and g to be differentiable at 0.The function g is obviously differentiable at 0, where g 0 (0) cos 0 1. By the definition of the derivative of f at 0: 2 xx x for x Qf (x) f (0)f (x) 0 x 0x 0for x R \ Q.x 0For any 0, let δ( ) . If 0 x 0 x δ( ), then f (x)/x x . It follows that f 0 (0) 0.Because f (0) g(0) 0 and g(x) 6 0 for x 6 0, we can use Theorem 6.3.1 to find the limit:f (x)f 0 (0)0 0 0.x 0 g(x)g (0)1limWe cannot avail ourselves of Theorem 6.3.3 because f is discontinuous (and therefore is not differentiable) everywhereon R except at 0. Let c R where c 6 0. Suppose (xn ) is a sequence that converges to c where xn Q for all n N.It follows that lim f (xn ) lim x2n c2 6 0. Now suppose (yn ) is a sequence that converges to c where yn R \ Q forall n N. We see that lim f (yn ) 0. If c is rational, (f (yn )) does not converge to f (c) c2 6 0. On the other hand, ifc is irrational, (f (xn )) does not converge to f (c) 0. By the Discontinuity Criterion (5.1.4), f is not continuous at anyc R where c 6 0.Problem 6.3-13. Show that if c 0, then limc cxc cxxx cc 1 ln c1 ln c .Page 4

Solution: Because c 0, let:f (x) xc cx xc ex ln c ,andg(x) xx cc ex ln x cc .Since f (c) g(c) 0, the desired limit is in 0/0 indeterminate form. The derivatives of f and g are:f 0 (x) cxc 1 (ln c)ex ln c ,andg 0 (x) ex ln x (1 ln c).For x 0, the derivatives of f and g exist and g 0 (x) 6 0. Accordingly, we may apply L’Hôpital’s Rule to find thedesired limit of f /g at c 0. We then have:f 0 (x)cxc 1 (ln c)ex ln cx c cx lim 0 limxcx c g (x)x cx c x cex ln x (1 ln c)lim cc cc ln cccc 1 (ln c)ec ln c ec ln c (1 ln c)cc cc ln c 1 ln c,1 ln cas we set out to show.Section 6.4Problem 6.4-3. Use induction to prove Leibniz’s rule for the nth derivative of a product:(f g)(n) n Xn (n k) (k)fg .kk 0Solution: The base case of n 1 is trivial. Using the product rule, we have:1 X1 (1 k) (k)fg .(f g) f g f g k000k 0Now we assume the inductive hypothesis holds for n 1, and we will show that the inductive hypothesis also holds forn 1. From the inductive hypothesis and the product rule, we have:" n #0X n(f g)(n 1) [(f g)(n) ]0 f (n k) g (k)kk 0n hiXn f (n 1 k) g (k) f (n k) g (k 1) .kk 0n n Xn (n 1 k) (k) X n (n k) (k 1) fg fg.kkk 0k 0By changing the index variable of the second sum to l k 1, we then have:(f g)(n 1 ) n Xnk 0 f (n 1) g f g (n 1) kf (n 1 k) g (k) n Xk 1n 1X l 1 nf (n 1 l) g (l) .l 1 n n (n 1 k) (k) Xnfg f (n 1 l) g (l)kl 1l 1Page 5

n Xn (n 1 k) (k)n(n 1 k) (k) fg fg fg fg.kk 1k 1 nPascal’s Rule states that k 1 nk n 1k . Applying this to the sum above, we find:(n 1)(n 1)(f g)(n 1) f (n 1) g f g (n 1) n Xn 1kk 1f (n 1 k) g (k)n 1X n 1 f (n 1 k) g (k) ,kk 0which is what we sought to show for n 1. Therefore, by the principle of mathematical induction, the inductive hypothesisholds for all n N.Problem 6.4-8. If f (x) ex , show that the remainder term in Taylor’s Theorem coverges to zero as n for each fixed x0and x.Solution: For any x0 and x, the remainder term of the nth Taylor polynomial is:Rn (x, x0 ) 1ec (x x0 )n 1 ,(n 1)!for some c between x and x0 . Let yn Rn (x, x0 ). For x 6 x0 , the quotient of yn 1 and yn is:yn 1 yn1c(n 2)! e (x1c(n 1)! e (x x0 )n 2 x0 )n 1 x x0.n 2For all n N, we have 0 1/(n 2) 1/n. By the Squeeze Theorem, lim 1/(n 2) 0 (x x0 ) lim 1/(n 2).Because this limit is less than one, it follows from Theorem 3.2.1 that lim yn 0. Therefore, as n , the remainderterm Rn (x, x0 ) 0. This shows that the Taylor polynomial of ex with an infinite number of terms equals ex .Problem 6.4-10. If x [0, 1] and n N, show that: nx2xn 1x3n 1 xln(1 x) x · · · ( 1).23nn 1Use this to approximate ln 1.5 with an error less than 0.01. Less than 0.001.Solution: Let f (x) ln(1 x). First, we will prove by induction that:f (n) (x) ( 1)n 1 (n 1)!.(1 x)nFor the base case n 1, we have f (1) (x) 1/(1 x) (( 1)0 0!)/(1 x)1 . Now we will assume the inductivehypothesis is true for n 1. For n 1, we have:f (n 1) (x) (f (n) )0 ( 1)n 1 (n 1)!( 1)n( 1)n n! .(1 x)n 1(1 x)n 1Therefore, the inductive hypothesis is true for all n N.The nth Taylor polynomial Pn (x) for x0 0 is:Pn (x) f (0) f 0 (0)x x 111 00f (0)x2 f 000 (0)x3 · · · f (n)2!3!n!x2x3xn · · · ( 1)n 1 .23nThe remainder term Rn (x) is:Page 6

Rn (x) ( 1)nxn 1,(n 1)(1 c)n 1for some c is between x and x0 . Because x [0, 1] by hypothesis and x0 0, it follows that c (0, 1) and:xn 1.(n 1)(1 c)n 1 Rn (x) By Taylor’s Theorem, ln(1 x) Pn (x) Rn (x), from which it follows that ln(1 x) Pn (x) Rn (x). Taking theabsolute value of both sides, we have: x2xn 1x3xnln(1 x) x . · · · ( 1)n 123n(n 1)(1 c)n 1Because c (0, 1), there is an upper bound on Rn (x) of xn 1 /(n 1). We infer: x2xn 1x3xnln(1 x) x · · · ( 1)n 1,23n(n 1)as we sought to show.We will now estimate ln 1.5. We see that ln 1.5 f (0.5). For an estimation error of less than 0.01, we must haveRn (0.5) 0.5n 1 /(n 1) 0.01. By trial and error, we find that R3 (0.5) 0.015625 and R4 (0.5) 0.00625.Consequently, we need at least a fourth-order Taylor polynomial to meet the estimation error. We find that P4 (0.5) 0.40104, which has an error within the sought tolerance of approximately 0.004425.We will now refine the estimate with an error less than 0.001. Again by trial and error, we find that R6 (0.5) 0.00116and R7 (0.5) 0.000458. A seventh-order Taylor polynomial will meet the estimation error. We see that P7 (0.5) 0.4058,which has an error of approximately 0.0003349.Problem 6.4-17. Suppose that I R is an open interval and that f 00 (x) 0 for all x I. If c I, show that the part of thegraph of f on I is never below to tangent line to the graph at (c, f (c)).Solution: Because f 0 (c) is the slope of the tangent line to f at c, we can express the tangent line as g(x) f (c) f 0 (c)(x c).The 1st Taylor polynomial P1 (x) is:P1 (x) f (c) f 0 (c)(x c),which is equal to g. The remainder term is:1 00f (d)(x c)2 ,2for some d between c and x. Because f 00 (d) 0 by hypothesis, R1 (x) 0 for all x I.From Taylor’s Theorem, if x I, then f (x) P1 (x) R1 (x) g(x) R1 (x). Since R1 (x) 0, necessarilyf (x) g(x), from which it follows that no point on I is below the tangent line to f at c.As an aside, note that f is convex on I by Theorem 6.4.6.R1 (x) Problem 6.4-15. Let f be continuous on [a, b] and assume the second derivative f 00 exists on (a, b). Suppose that the graphof f and the line segment joining the points (a, f (a)) and (b, f (b)) intersect at a point (x0 , f (x0 )) where a x0 b. Showthat there exists a point c (a, b) such that f 00 (c) 0.Solution: Let g be the line segment between pa (a, f (a)) and pb (b, f (b)), where g(x) g(a) m(x a) andm (g(b) g(a))/(b a). Because px0 (x0 , f (x0 )) is on the line segment given by g, the line segment between pa andpx0 and the line segment between px0 and pb each have slope m. By the Intermediate Value Theorem, there is a pointd1 (a, x0 ) and a point d2 (x0 , b) such that:f 0 (d1 ) f (x0 ) f (a) m,x0 aPage 7

andf 0 (d2 ) f (b) f (x0 ) m.b x0Let φ : (a, b) R where φ(x) f 0 (x) m. Because f 00 exists everywhere on (a, b), by Theorem 6.1.2 f 0 (and thereforeφ) is necessarily continuous and differentiable on (a, b). Given that φ(d1 ) φ(d2 ) 0, by Rolle’s Theorem there is a pointc (d1 , d2 ) such that φ0 (c) 0. Since φ0 (x) f 00 (x), it follows that f 00 (c) φ0 (c) 0. Because (d1 , d2 ) (a, b), thereis a point c on (a, b) where f 00 (c) 0.Problem 6.4-16. Let I R be an open interval, let f : I R be differentiable on I, and suppose f 00 (a) exists at a I.Show thatf (a h) 2f (a) f (a h)h2Give an example where this limit exists but the function does not have a second derivative at a.f 00 (a) limh 0Solution: The second-order Taylor polynomial for f at x0 a is:1P2 (x) f (a) f 0 (a)(x a) f 00 (a)(x a)2 .2Applying Taylor’s Theorem for f (a h) and f (a h), we have:11f (a h) P2 (a h) R2 (a h) f (a) f 0 (a)h f 00 (a)h2 f 000 (d1 )h3 ,26(7)for some d1 (a, a h) and:11f (a h) f (a) f 0 (a)h f 00 (a)h2 f 000 (d2 )h3 ,26(8)for some d2 (a h, a).Adding (7) and (8), we have:11f (a h) f (a h) 2f (a) f 00 (a)h2 f 000 (d1 )h3 f 000 (d2 )h3 .66from which we have:1f (a h) 2f (a) f (a h)1.f 00 (a) f 000 (d1 )h f 000 (d2 )h 66h2Taking the limit of both sides as h 0 achieves the desired result:f 00 (a) limh 0f (a h) 2f (a) f (a h).h2Now we will turn to an example where this limit exists for a function, but that function has no second derivative.Specifically, this limit exists for f (x) signum(x) at a 0, where:signum(0 h) 2(0) signum(0 h)0 lim 2 0.2h 0h 0 hhf 00 (0) limClearly the limit exists, but we know that f is not differentiable at x 0 because it is not continuous at that point.Therefore f cannot have a second derivative.Problem 6.4-18. Let I R be an interval and let c I. Suppose that f and g are defined on I and that the derivativesf (n) , g (n) exist and are continuous on I. If f (k) (c) 0 and g (k) (c) 0 for k 0, 1, ., n 1, but g (n) (c) 6 0, show thatf (x)f (n) (c) (n) .x c g(x)g (c)limPage 8

Solution: There is a straightforward proof involving Taylor’s Theorem, using the remainder of f and g to show the limit.To get some practice with proving the existence of neighborhoods with certain properties, I will prove the proposition usingL’Hôpital’s Rule.Because f (c) g(c) 0, the desired limit is in 0/0 indeterminate form. Applying L’Hôpital’s Rule iteratively—if wecan—will clearly result in the desired solution. To avail ourselves of the rule, however, we must show that for each g (k) ,there is a neighborhood of c on which g (k) exists and is zero only at c.First, we will prove by induction for l N where l n, if g (n l) (c) 0 (which is the case by hypothesis), there is aneighborhood Ul of c on which g (n 1) is differentiable and, if x Ul and x 6 c, then g (n 1) (x) 6 0. Let L g (n) (c),which must be non-zero by hypothesis. For l 1, suppose L . Because g (n) exists at c, there is a δ1 0 such that ifx I and 0 x c δ1 , then:g (n 1) (x)g (n 1) (x) g (n 1) (c) L L .x cx cIf we assume g (n 1) (x) 0, theng (n 1) (x) L L L ,x cwhich contradicts the requirement that this value must be less than . Therefore, g (n 1) (x) 6 0. We conclude that thereis a δ1 -neighborhood U1 of c such that if x U1 \{c}, then g (n 1) (x) 6 0.Assume that the inductive hypothesis holds for l. If l 1 n, then by the inductive hypothesis, there is a Ul of c suchthat if x Ul \{c}, then g (n l) (x) 6 0. Because g (n l) is the derivative of g (n l 1) , we infer that g (n l 1

Bartle - Introduction to Real Analysis - Chapter 6 Solutions Section 6.2 Problem 6.2-4. Let a 1;a 2;:::;a nbe real numbers and let fbe de ned on R by f(x) Xn i 0 (a i x)2 forx2R: Find the unique point of relative minimum for f. Solution: The rst derivative of fis: f0(x) 2 Xn i 1 (a i x): Equating f0to zero, we nd the relative extrema c2R as follows: f0(c) 2 Xn i 1 (a i c) 2 " nc Xn i .